Irreducibility
Definition A topological space is called reducible if
it can be written as the union of two proper nonempty closed
subsets. A space that is not reducible is called
irreducible.
Proposition An algebraic set X is irreducible if and
only if the ideal I(X) is prime.
Proof: If
I(X) is not prime, then there exist
polynomials
f,
g∉ I(X)
with product
fg∈ I(X). So, we can write
X = (X∩ Z(f)) ∪ (X∩
Z(g))
as a nontrivial decomposition of
X.
Conversely, suppose
X= X1∪ X2 is a
decomposition of
X as a union of nonempty proper
closed subsets. Then each ideal
I(Xi) contains
I(X) properly. So, we can choose functions
fi∈ I(Xi)\ I(X). Then the
product
f1f2∈ I(X), so the ideal
I(X) is not prime.
Corollary There is a one-to-one correspondence between irreducible
algebraic sets in An and prime ideals in
k[x1, …, xn].
Theorem
Let X be an algebraic set in
An. Then there exists a unique collection of
irreducible algebraic sets
X1, …, Xm such
that
- X= X1∪ X2∪…∪
Xm, and
- Xi⊂ Xj if and only if
i=j.
Proof: (Existence) Suppose the theorem does not hold for an algebraic
set
X. In particular,
X must be
reducible. Write
X=X1∪ Y1. The theorem must
also fail for at least one of the pieces in this decomposition; say,
X1. Now write
X1=X2∪
Y2 and repeat. In this way, one constructs an infinite
decreasing chain of algebraic sets
X ⊃ X1 ⊃ X2 ⊃
….
However, the chain of associated ideals
I(X) ⊂ I(X1) ⊂ I(X2) ⊂
…
must terminate, by Hilbert's Basis Theorem. Therefore,
X has a finite decomposition as a union of
irreducibles. The condition prohibiting containments can be realized
by throwing out any redundant items in the decomposition.
(Uniqueness) Suppose
X=∪iXi = ∪jYj
are two irredundant decompositions of
X as a union
of irreducible algebraic sets. For each
i, we can
write
Xi = Xi∩ X = Xi∩(∪jYj) =
∪j(Xi∩ Yj).
Because
Xi is irreducible, there exists some
j with the property that
Xi = Xi∩
Yj ⊂ Yj. By interchanging the roles of
the decompositions, there exists some
i' with
Xi ⊂ Yj ⊂ Xi'.
Because the decompositions are irredundant, this last chain of
inclusions must actually consist of equalities, and the result
follows.
Corollary Let
f∈ k[x1,…,xn]. Write
f=f1n1… fsns
as a product of irreducible factors. Then
Z(f) = Z(f1) ∪…∪ Z(fs)
is the unique decomposition of its zero set into irreducible
components. Furthermore,
I(Z(f)) = (f1f2… fs).
Proof: The function f1 must vanish on some
component X1 of X. But then
X1⊂ Z(f1), which forces X1 =
Z(f1)∩ X.
Definition An algebraic set defined by one polynomial in
An is called a
hypersurface. The irreducible hypersurfaces are in
one-to-one correspondence with the irreducible polynomials.
Theorem The product of two irreducible algebraic sets is irreducible.
Proof: Let
X and
Y be
irreducible, and suppose that
X× Y = Z1∪
Z2 is a decomposition of the product. For each point
x∈ X, we have
Y = x× Y = ((x× Y)∩ Z1)∪
((x× Y)∩ Z2).
So, at least one of the factors on the right contains this copy of the
irreducible set
Y. Now define
Xi⊂
X to be the set of all
x∈ X such
that
x× Y ⊂ Zi. We have already
observed that
X=X1∪ X2. However,
X is also irreducible and the
Xi are closed subsets. Therefore, we must have
X=X1 and
X× Y = Z1.
Definition An affine algebraic variety over an
algebraically closed field k is an irreducible
algebriac set in An with the induced topology
and the (reduced) induced structure.
Examples
- An is a variety.
- Among the curves of the first section, only (v), (x), and
(xi) are not varieties.
- A point is a variety; two points are not.
Comments on this web site should be addressed to the
author:
Kevin R. Coombes
Department of Biomedical Informatics
The Ohio State University
Columbus, Ohio 43210